Suppose that $$D=\{z:0<x<a,0<y<b\}$$ and that $$D'=\{w:0<u<c,0,v<d\}$$ Then there is a conformal mapping $f$ of $D$ onto $D'$ whose homeomorphic extension $\tilde{f}$ to $\overline{D}$ satisfies $$\tilde{f}(0)=0, \tilde{f}(a)=c,\tilde{f}(ib)=id,$$ and $$\tilde{f}(a+ib)=c+id$$ if and only if $$\frac{a}{b}=\frac{c}{d}$$
The hint given in the text is to show that $f$ can be extended to a conformal self mapping of the complex plane. By using the Schwarz reflection principle, I can easily "reflect" over the boundaries of the rectangle to cover the whole complex plane but how does that help me in proving the claim?
Any help would be greatly appreciated! Thanks.
Every conformal self-map of the whole plane is of the form $\alpha z +\beta$.